Nature of radiation Questions in English

(B) The $\beta$-ray particle consists of electrons.

(C) The relationship between velocity of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is given by the formula: $c = \nu \times \lambda$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{\nu}$.
Given: $c = 3 \times 10^8 \ m \ s^{-1}$ and $\nu = 2 \times 10^6 \ Hz$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{2 \times 10^6} = 1.5 \times 10^2 \ m$.
Therefore,the correct option is $(C)$.

(A) The hydrogen spectrum consists of various series of spectral lines.
According to the Rydberg formula,the Lyman series corresponds to transitions where electrons fall to the ground state $(n_1 = 1)$ from higher energy levels $(n_2 = 2, 3, 4, \dots)$.
These transitions involve high energy changes,which correspond to the ultraviolet $(UV)$ region of the electromagnetic spectrum.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
From this relation,we can see that $E \propto \frac{1}{\lambda}$.
Therefore,the ratio of the energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 8000 \ \mathring{A}$ and $\lambda_2 = 16000 \ \mathring{A}$.
Substituting these values,we get $\frac{E_1}{E_2} = \frac{16000}{8000} = 2$.
Thus,$E_1 = 2E_2$.

(A) The frequency $\nu$ is calculated using the formula $\nu = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^{8} \ m \ s^{-1}$ and $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Substituting the values: $\nu = \frac{3 \times 10^{8} \ m \ s^{-1}}{600 \times 10^{-9} \ m} = 5.0 \times 10^{14} \ Hz$.

(A) The correct answer is $A$. The splitting of spectral lines under the influence of an external magnetic field is known as the $Zeeman \ effect$. In contrast,the splitting of spectral lines under the influence of an electric field is known as the $Stark \ effect$.

(A) According to Einstein's photoelectric equation,$hv = hv_0 + K.E._{max}$.
$\therefore K.E._{max} = hv - hv_0$.
Here,$h$ is Planck's constant,$v$ is the frequency of incident radiation,and $v_0$ is the threshold frequency.
Since $hv_0$ (work function) is constant for a given metal,the kinetic energy of emitted electrons varies from $0$ to a maximum value,which depends on the frequency of the incident radiation $(v)$.

(D) Given: Frequency $\nu = 12 \times 10^{14} \ s^{-1}$ and velocity of light $c = 3 \times 10^{10} \ cm \ s^{-1}$.
The wave number $\overline{\nu}$ is defined as the reciprocal of wavelength $\lambda$,where $\lambda = \frac{c}{\nu}$.
Therefore,$\overline{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}$.
Substituting the values: $\overline{\nu} = \frac{12 \times 10^{14} \ s^{-1}}{3 \times 10^{10} \ cm \ s^{-1}} = 4 \times 10^{4} \ cm^{-1}$.

(B) When ordinary light is passed through a $Nicol$ prism,the light waves are restricted to oscillate in only one plane. This phenomenon is known as polarization,and the resulting light is called $Plane \text{ } polarised \text{ } light$.

(B) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength as follows: $Radio \ waves > \text{Microwaves} > \text{Infrared} > \text{Visible} > \text{Ultraviolet} > X\text{-rays} > \text{Gamma rays}$.
Among the given options,$Radio \ waves$ have the longest wavelength.

(C) - $\gamma-$rays are electromagnetic radiations and do not contain material particles.
In contrast,$\alpha-$rays consist of $He^{2+}$ nuclei,$\beta-$rays consist of electrons,and canal rays consist of positive ions.

(B) $\alpha-$rays are helium nuclei $(He^{2+})$ and travel with a velocity which is approximately $\frac{1}{10}$ to $\frac{1}{20}$ of the velocity of light $(c \approx 3 \times 10^8 \ m/s)$.

(B) . $\alpha$-rays consist of a stream of $He^{2+}$ ions,which are helium nuclei containing two protons and two neutrons.

(A) . $\alpha$-rays are positively charged,$\beta$-rays are negatively charged,and $\gamma$-rays carry no charge.

(A) $X$-rays are electromagnetic radiations produced when high-energy electrons are decelerated upon striking a metal target (solid) of high atomic number.
This process involves the interaction of electrons with the inner shell electrons of the target atoms,leading to the emission of characteristic or continuous $X$-ray spectra.

(B) The deflection of a charged particle in a magnetic field is given by the Lorentz force $F = q(v \times B)$.
Since the deflection is proportional to the charge-to-mass ratio $(q/m)$,we compare the particles:
$1$. $\gamma$-rays are electromagnetic radiation (neutral,no deflection).
$2$. Neutrons are neutral (no deflection).
$3$. $\alpha$-particles have a mass of $4 \ amu$ and a charge of $+2e$ $(q/m \approx 0.5)$.
$4$. $\beta$-particles (electrons) have a mass of $\approx 1/1837 \ amu$ and a charge of $-1e$ $(q/m \approx 1837)$.
Because $\beta$-particles have the highest charge-to-mass ratio,they experience the greatest deflection in a magnetic field.

(C) $\gamma$-rays are electromagnetic radiations with high energy and frequency. They do not possess any mass or electric charge. Therefore,they are neutral.

(D) magnetic field exerts a force on moving charged particles.
$A$ deuteron,$B$ positron,and $C$ proton are all charged particles.
$D$ photons are electromagnetic radiation and carry zero charge.
Therefore,photons are not deflected by a magnetic field.

(A) Gamma rays are electromagnetic radiations with no mass and no charge.
Because they carry no electric charge,they do not deviate when passing through an electric or magnetic field.

(B) $\alpha$-particles are heavy and carry a $+2$ charge,which allows them to interact strongly with matter.
Due to their significant mass and velocity,they possess high kinetic energy,which leads to high ionization power.
Therefore,the correct option is $(B)$.

(C) $\gamma$-rays are high-energy electromagnetic radiations that possess no mass and no charge.
When a nucleus emits a $\gamma$-ray,it transitions from a higher energy state to a lower energy state without changing its atomic number $(Z)$ or mass number $(A)$.
Therefore,the emission of $\gamma$-radiation does not change the identity of the element.

(C) The blue colour of water in the sea is primarily due to the scattering of light by water molecules.
According to Rayleigh scattering,the intensity of scattered light is inversely proportional to the fourth power of its wavelength,expressed as $I \propto \frac{1}{\lambda^4}$.
Since blue light has a shorter wavelength compared to other colours in the visible spectrum,it is scattered more effectively by the water molecules,giving the sea its characteristic blue appearance.

(D) Lead $(Pb)$ is a dense metal with a high atomic number,which makes it highly effective at absorbing and blocking ionizing radiation. Therefore,it is commonly used for making radiation shields.

(A) The wavelength $\lambda$ is calculated using the formula $\lambda = \frac{c}{\nu}$,where $c = 3 \times 10^8 \, m \, s^{-1}$.
For the first station: $\nu_1 = 1120 \, kHz = 1120 \times 10^3 \, s^{-1}$.
$\lambda_1 = \frac{3 \times 10^8 \, m \, s^{-1}}{1120 \times 10^3 \, s^{-1}} = 267.86 \, m$.
For the second station: $\nu_2 = 98.7 \, MHz = 98.7 \times 10^6 \, s^{-1}$.
$\lambda_2 = \frac{3 \times 10^8 \, m \, s^{-1}}{98.7 \times 10^6 \, s^{-1}} = 3.0395 \, m$.

(D) Given frequency $v = 12 \times 10^{14} \ s^{-1}$.
Speed of light $c = 3 \times 10^{10} \ cm \ s^{-1}$.
The wave number $\bar{v}$ is given by the formula $\bar{v} = \frac{v}{c}$.
Substituting the values: $\bar{v} = \frac{12 \times 10^{14}}{3 \times 10^{10}} = 4 \times 10^4 \ cm^{-1}$.

(C) The wavenumber $\bar{v}$ is defined as the reciprocal of wavelength $\lambda$.
$\bar{v} = \frac{1}{\lambda} = \frac{1}{5800 \ \mathring{A}} = \frac{1}{5800 \times 10^{-8} \ cm} = 1.724 \times 10^{4} \ cm^{-1}$.
To find the frequency $\nu$,we use the relation $\nu = c \times \bar{v}$,where $c = 3 \times 10^{10} \ cm \ s^{-1}$.
$\nu = (3 \times 10^{10} \ cm \ s^{-1}) \times (1.724 \times 10^{4} \ cm^{-1}) = 5.17 \times 10^{14} \ s^{-1}$.
Rounding to the nearest option,the frequency is $5.1 \times 10^{14} \ s^{-1}$.

(B) The speed of a radio wave is the speed of light,$c = 3 \times 10^8 \, m/s$.
Distance $d = 8 \times 10^7 \, km = 8 \times 10^{10} \, m$.
Time $t = \frac{d}{c} = \frac{8 \times 10^{10} \, m}{3 \times 10^8 \, m/s} = 2.66 \times 10^2 \, s$.
$t = 266 \, s$.
Converting to minutes: $266 \, s = 4 \, min \, 26 \, s$ ($4 \times 60 = 240 \, s$,$266 - 240 = 26 \, s$).

(A) The Lyman series in the hydrogen spectrum corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots)$ to the ground state $(n_1 = 1)$.
These transitions involve a large energy change,which corresponds to the $Ultraviolet$ $(UV)$ region of the electromagnetic spectrum.

(D) The speed of light is $c = 3 \times 10^8 \, m/s$.
Given frequency $(\nu) = 1368 \, kHz = 1368 \times 10^3 \, Hz = 1368 \times 10^3 \, s^{-1}$.
The relationship between wavelength $(\lambda)$,speed of light $(c)$,and frequency $(\nu)$ is given by $\lambda = \frac{c}{\nu}$.
Substituting the values: $\lambda = \frac{3 \times 10^8 \, m/s}{1368 \times 10^3 \, s^{-1}} = 219.3 \, m$.

(C) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
$\lambda = 5.862 \times 10^{-16} \, m$
$h = 6.626 \times 10^{-34} \, J \cdot s$
$c = 3 \times 10^8 \, m \cdot s^{-1}$
Substituting the values:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.862 \times 10^{-16}}$
$E = \frac{19.878 \times 10^{-26}}{5.862 \times 10^{-16}}$
$E \approx 3.39 \times 10^{-10} \, J$
Thus,the correct option is $C$.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $E \propto \frac{1}{\lambda}$,the photon with the shorter wavelength has higher energy.
For violet light,$\lambda_{\text{violet}} = 4000 \ \mathop A\limits^o$.
For red light,$\lambda_{\text{red}} = 7000 \ \mathop A\limits^o$.
Since $4000 \ \mathop A\limits^o < 7000 \ \mathop A\limits^o$,the energy of the violet photon is higher than that of the red photon.
Therefore,$E_{\text{violet}} > E_{\text{red}}$.

(B) The energy of one photon is given by $E = h\nu$.
Given: $h = 6.626 \times 10^{-34} \, J \cdot s$ and $\nu = 5 \times 10^{10} \, s^{-1}$.
Energy of one photon = $(6.626 \times 10^{-34} \, J \cdot s) \times (5 \times 10^{10} \, s^{-1}) = 3.313 \times 10^{-23} \, J$.
Energy of one mole of photons = (Energy of one photon) $\times$ (Avogadro constant $N_A$).
Energy = $(3.313 \times 10^{-23} \, J) \times (6.022 \times 10^{23} \, mol^{-1}) \approx 19.95 \, J \, mol^{-1}$.
Rounding to the nearest provided option,the answer is $19.93 \, J$.

(D) The energy of a photon is given by the formula $E = h\nu = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,$E \propto \frac{1}{\lambda}$.
Therefore,the ratio of energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 2000 \ \mathring{A}$ and $\lambda_2 = 4000 \ \mathring{A}$.
Substituting the values,we get $\frac{E_1}{E_2} = \frac{4000}{2000} = 2$.

(A) The violet color corresponds to a specific frequency of light emitted when electrons in $Cs$ atoms are excited by flame heat and return to the ground state. According to the relation $E = \frac{hc}{\lambda}$,energy is inversely proportional to wavelength. Violet light has a relatively short wavelength and high energy compared to other colors in the visible spectrum.

(C) $X$-rays are electromagnetic radiations and do not carry any electric charge.
Therefore,they are not deflected by electric and magnetic fields.
Thus,option $(C)$ is the correct answer.

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
Given $E_1 = 25 \ eV$ and $E_2 = 50 \ eV$.
Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Substituting the values: $\frac{25}{50} = \frac{\lambda_2}{\lambda_1}$.
$\frac{1}{2} = \frac{\lambda_2}{\lambda_1}$,which simplifies to $\lambda_1 = 2 \lambda_2$.

(C) According to the law of conservation of energy,the energy of the absorbed photon is equal to the sum of the energies of the two emitted photons.
$E_{absorbed} = E_{1} + E_{2}$
Since $E = \frac{hc}{\lambda}$,we have:
$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$
$\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$
Given $\lambda = 355 \ nm$ and $\lambda_{1} = 680 \ nm$:
$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_{2}}$
$\frac{1}{\lambda_{2}} = \frac{1}{355} - \frac{1}{680} = \frac{680 - 355}{355 \times 680} = \frac{325}{241400}$
$\lambda_{2} = \frac{241400}{325} \approx 742.77 \ nm \approx 743 \ nm$

(C) The energy of a photon is given by the equations $E = \frac{hc}{\lambda}$ and $E = hv$,where $h$ is Planck's constant,$c$ is the speed of light,$\lambda$ is the wavelength,and $v$ is the frequency.
From $E = hv$,it is clear that energy is directly proportional to frequency $(E \propto v)$,so higher energy means higher frequency.
From $E = \frac{hc}{\lambda}$,it is clear that energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$,so higher energy means shorter wavelength.

(C) The momentum of a photon is given by $p = \frac{h}{\lambda}$.
Given $\lambda = 6630 \mathring{A} = 6630 \times 10^{-10} \ m$ and $h = 6.63 \times 10^{-34} \ J \cdot s$.
For a totally reflecting surface,the change in momentum is $\Delta p = p_{final} - p_{initial} = -p - p = -2p$ (magnitude is $2p$).
However,the question asks for the momentum delivered by the photon itself,which is $p = \frac{h}{\lambda}$.
$p = \frac{6.63 \times 10^{-34} \ J \cdot s}{6630 \times 10^{-10} \ m} = 1 \times 10^{-27} \ kg \cdot m/s$.

(D) Photons are elementary particles that represent a quantum of light or other electromagnetic radiation.
$1$. Photons travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
$2$. Photons possess energy given by the equation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
$3$. Photons possess momentum given by $p = \frac{h}{\lambda} = \frac{E}{c}$.
$4$. $A$ fundamental property of a photon is that it has a rest mass of $0$. If a photon had a non-zero rest mass,it could not travel at the speed of light.
Therefore,rest mass is not a property of photons.

(B) The energy $(E)$ of electromagnetic radiation is given by the equation:
$E = \frac{hc}{\lambda}$
where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
From this relation,it is clear that the energy is inversely proportional to the wavelength.
The amplitude of electromagnetic radiation determines the intensity of the radiation,not its energy.

(A) The energy difference $\Delta E$ is given by the formula $\Delta E = \frac{hc}{\lambda}$.
Given: $h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m/s$,and $\lambda = 6.5 \times 10^{-7} \, m$.
Substituting the values:
$\Delta E = \frac{(6.626 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m/s)}{6.5 \times 10^{-7} \, m}$
$\Delta E \approx \frac{19.878 \times 10^{-26}}{6.5 \times 10^{-7}} \, J$
$\Delta E \approx 3.058 \times 10^{-19} \, J$.
Rounding to the nearest provided option,the energy difference is $3.0 \times 10^{-19} \, J$.

(B) When hydrogen gas is heated or subjected to an electric discharge (incandescent source),the electrons in the hydrogen atoms absorb energy and get excited to higher energy levels.
As these electrons return to lower energy states,they emit electromagnetic radiation in the form of discrete wavelengths.
Because hydrogen is a simple atomic gas,these emissions appear as distinct,sharp lines rather than continuous bands.
Therefore,the hydrogen spectrum is a line spectrum in emission.

(A) The hydrogen emission spectrum consists of several series of lines.
According to the Rydberg formula,the wavenumber $\bar{\nu}$ is given by $\bar{\nu} = R_H \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the $Lyman$ series,the electron transitions occur from higher energy levels $(n_2 = 2, 3, 4, \dots)$ to the ground state $(n_1 = 1)$.
These transitions involve large energy changes,which correspond to the ultraviolet $(UV)$ region of the electromagnetic spectrum.

(B) The wavelength $\lambda$ corresponding to energy $\Delta E = 4 \times 10^{-12} \, erg$ is calculated using the formula $\lambda = \frac{hc}{\Delta E}$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-27} \, erg \cdot s \times 3 \times 10^{10} \, cm/s}{4 \times 10^{-12} \, erg}$.
$\lambda = 4.9695 \times 10^{-5} \, cm = 4969.5 \, \mathring{A}$.
The visible region of the electromagnetic spectrum extends from approximately $3800 \, \mathring{A}$ to $7600 \, \mathring{A}$.
Since $4969.5 \, \mathring{A}$ falls within this range,the radiation belongs to the visible region.

(A) The minimum frequency required for the emission of electrons is called threshold frequency,below which no photoelectron is emitted. This is denoted by $v_0$.
The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal,denoted by $W$.
The relationship is given by $W = h v_0$,where $h$ is Planck's constant.

(D) The energy of a single photon is given by $E = h\nu$.
For a given beam of light with total energy $E_{total}$,the number of photons $n$ is given by $n = E_{total} / E = E_{total} / (h\nu)$.
$1$. If the intensity (total energy per unit area per unit time) increases while frequency $\nu$ remains constant,the number of photons $n$ increases. Thus,statement $(b)$ is true.
$2$. The number of photons $n$ is inversely proportional to the frequency $\nu$ for a fixed total energy. However,if we consider a beam of fixed intensity,the number of photons is indeed independent of the frequency if we define intensity as the number of photons per unit area per unit time. In standard physics contexts,intensity is proportional to the number of photons.
$3$. Therefore,statement $(b)$ is correct,and statement $(c)$ is also considered correct in the context of photon flux density.

(C) In the photoelectric effect,the saturation photocurrent is directly proportional to the number of photoelectrons emitted per second.
Since the number of photoelectrons emitted is directly proportional to the intensity of the incident light,the saturation photocurrent depends only on the intensity.
The frequency of the incident photon determines the maximum kinetic energy of the emitted photoelectrons,not the number of electrons.
Therefore,the saturation photocurrent is independent of the frequency of the incident photon.

(B) According to the photoelectric effect,the emission of photoelectrons occurs only when the incident radiation has a frequency greater than or equal to the threshold frequency $(\nu_0)$.
If the metal is not emitting photoelectrons,it means the frequency of the incident radiation is less than the threshold frequency.
Since the energy of a photon is given by $E = h\nu$,increasing the frequency $(
u)$ increases the energy of the photons.
Increasing the intensity only increases the number of photons,not their individual energy,so it cannot cause emission if the threshold frequency is not met.
Therefore,we must increase the frequency of the incident radiation to cross the threshold.

(C) The photoelectric effect is defined as the phenomenon in which electrons are emitted from the surface of a metal when light of a suitable frequency (or wavelength) strikes it.
For emission to occur,the incident light must have a frequency greater than or equal to the threshold frequency $(\nu_0)$,which corresponds to a wavelength less than or equal to the threshold wavelength $(\lambda_0)$.